Re: Identify the discontinuity. f'(x) for the f(x).

Part d is a v shape Just the top part of your graph. Ofcourse, you don't know what the value of 'a' is so you don't know if the vertex is above or below the x axis, but you do know it is a V
When x>0 the gradient is +1, when x<0 the gradient is -1
What happens when x=0? The graph is not defined. There is no f(x) value when x=0, f(x) is undefined when x=0 so f(x) is discontinuous when x=0
You show this on a graph by putting an open circle around this point (0,a)

Now when x>0 f'(x)=1 This is just a line horizontal to the x axis starting at (0,1) x=0 is not included so it will start with an open circle
And, when x<0 f'(x)=-1 This is just a line horizontal to the x axis starting at (0,-1) x=0 is not included so it will start with an open circle

So f'(x) is discontinuous at x=0 and the two halves start in very different places.

(Sorry, I don't know what the words are for different types of discontinuity)